Homeschool Guide: These lesson plans are a guide for parents. Content may contain errors — always cross-reference with official exam board specifications.

graphs and rates of change

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4 detailed 50-minute lessons with teaching scripts, worked examples, parent guides, and assessment criteria.

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Lesson Overview

Total Lessons: 4
Tier: Foundation and Higher
Duration: 50 minutes per lesson (200 minutes total)
Exam Boards: AQA, Edexcel, OCR, Eduqas, CCEA

Learning Objectives

Prerequisites

Materials & Equipment

Lesson 1: Introduction: graphs and rates of change

Duration: 50 minutes

Starter Activity (5 minutes)

Quick Recall

Write down everything you already know about graphs and rates of change. Then check against the key terms: Rate of Change. Use a mini-whiteboard or paper.

Main Content (35 minutes)

Parent/Teacher Guide:
Before lesson: Read the script below. Pre-teach key vocab: Rate of Change.
If stuck: Re-read the revision notes (link above), then break the content into smaller steps.
Extension: See the Stretch & Challenge ideas in Lesson 4.
Teaching Script (35 mins):
Mins 0-5 - Hook: "Today: graphs and rates of change. By the end you will be able to answer exam questions on it unaided. It connects to the rest of Mathematics because the ideas here recur across the spec."
Mins 5-20 - Direct Instruction: Work through the core ideas below one at a time; after each, ask your student to explain it back in their own words.
Mins 20-30 - Guided Practice: Model the worked example together, then let your student attempt the first practice question with guidance.
Mins 30-35 - Independent Practice: 2-3 practice questions from Lesson 3 below, with immediate feedback.
First Look

Start with the revision notes summary, then attempt: Find the gradient of the line through (3, 8) and (7, 24).

Plenary (5 minutes)

Check Out

Your student states one thing they learned and one question they still have about graphs and rates of change.

Lesson 2: Core Concepts: graphs and rates of change

Duration: 50 minutes

Starter Activity (5 minutes)

Review Previous Lesson

Quick recap: write 3 key points from Lesson 1 on graphs and rates of change. Check them against the notes below.

Main Content (35 minutes)

Rate of Change: The gradient of a straight line graph represents the rate of change of y with respect to x. It tells us how much y changes for each unit change in x.
TermMeaningExample
Distance-TimeDistanceTime
Velocity-TimeVelocityTime
Cost-QuantityCostQuantity
Height-AgeHeightAge

Practice (10 minutes)

Q: Find the gradient of the line through (3, 8) and (7, 24).

Answer: 4

Plenary (5 minutes)

Explain Back

Your student teaches the key points back to you without looking. Fill any gaps immediately.

Lesson 3: Application: graphs and rates of change

Duration: 50 minutes

Starter Activity (5 minutes)

Quick Recall

Recall the key terms: Rate of Change. Define each in one sentence.

Main Content (35 minutes)

Parent/Teacher Guide: Let your student attempt each question alone first, then compare with the model answer. Award method marks for correct working even if the final answer is wrong.

Q1: Find the gradient of the line through (3, 8) and (7, 24).

Answer: 4

Q2: A distance-time graph shows points (0, 0) and (3, 180). Distance in miles, time in hours. Find the speed.

Answer: 60 mph

Q3: A velocity-time graph has gradient -4. What does this mean?

Answer: Deceleration of 4 m/s² (slowing down)

Q4: Cost of fuel: graph passes through (10, 14) and (25, 35). Units: litres and £. Find the cost per litre.

Answer: £1.40 per litre

Q5: A car travels 150 miles in 3 hours at constant speed. What is the gradient of the distance-time graph?

Answer: 50 (miles per hour)

Plenary (5 minutes)

Error Review

Review any questions answered incorrectly. Identify whether the error was knowledge, method, or reading the question.

Lesson 4: Exam Practice: graphs and rates of change

Duration: 50 minutes

Starter Activity (5 minutes)

Command Words

Review what these command words require: state (one point), describe (say what happens), explain (say why), compare (both sides), evaluate (judgement).

Main Content (35 minutes)

Extended Answer

Extended question: Extended Answer 6 marks: A car accelerates uniformly from rest, reaching a speed of v m/s in 8 seconds. It then travels at this constant speed for 12 seconds before decelerating uniformly to rest in 4 seconds. The total distance travelled is 560 m. (a) Sketch the speed-time graph. (b) Find the value of v. (c) Calculate the acceleration and deceleration. <div class="

(a) Graph: triangle (0 to 8s), rectangle (8 to 20s), triangle (20 to 24s). (b) Total area = ½ × 8 × v + 12 × v + ½ × 4 × v = 4v + 12v + 2v = 18v. 18v = 560, so v = 560/18 = 31.1 m/s (1 d.p.). (c) Acceleration = 31.1/8 = 3.89 m/s². Deceleration = 31.1/4 = 7.78 m/s². Mark scheme: M1 sketch with correct shape, A1 labels, M1 total area expression, A1 v = 31.1, M1 a = v/t, A1 both values

Exam Tips: Gradient = (change in y) / (change in x) | Distance-time: gradient = speed | Velocity-time: gradient = acceleration | Always state units in your answer | Read axes carefully to identify what the gradient represents
Common Errors: Watch Out! 1. Wrong: Reading the speed from a distance-time graph directly off the y-axis at a point Correct: Speed = gradient of the distance-time graph, not the value on the axis. A flat line means zero speed regardless of the distance value. 2. Wrong: Calculating the area under a distance-time graph to find distance Correct: Area under a speed-time graph gives distance. Area under a distance-time graph has no useful physical meaning. 3. Wrong: Forgetting that a negative gradient on a speed-time graph means deceleration, not going backwards Correct: A negative gradient on a speed-time graph means the object is slowing down. The object only reverses when speed goes negative.
AO3 - Reasoning & Interpretation: Reasoning and Interpretation Two cyclists A and B race along the same 10 km route. Cyclist A starts from rest, accelerates to 20 km/h in 2 minutes, then cycles at 20 km/h for the rest. Cyclist B cycles at a steady 18 km/h from the start. (a) Who completes the 10 km first? (b) A student says "A's average speed must be more than 18 km/h since A reaches 20 km/h." Is this necessarily true? (c) What assumption have we made about A's acceleration phase? Answers: (a) A: distance during acceleration ≈ ½ × 20 × (2/60) = 0.333 km. Remaining: 9.667 km at 20 km/h = 0.483 hr ≈ 29 min. Total ≈ 31 min. B: 10/18 = 0.556 hr ≈ 33.3 min. A wins. (b) Not necessarily — A's average speed depends on the full journ
Stretch & Challenge (Grade 8-9):
  • Synoptic links: explain how graphs and rates of change connects to another Mathematics topic you have studied
  • Real-world: research one real-world use or example of graphs and rates of change
  • Critical: "What are the limitations of the models used in graphs and rates of change?"

Plenary (5 minutes)

Assessment Criteria
  • Got it: Confident explanation + correct worked examples
  • Getting there: Main points OK, needs support with detail
  • Not yet: Confused on key concepts - re-run Lesson 2

Homework & Consolidation

Recommended Resources

🎓 Smart Lesson (Guided)